A pulley of radius $2$ m is rotated about its axis by a force $F = (20t - 5t^2)$ N (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10\ \text{kg m}^2$, the number of rotations made by the pulley before its direction of motion is reversed, is
Answer: (B) more than $3$ but less than $6$
$$\alpha = \frac{FR}{I} = \frac{2(20t - 5t^2)}{10} = 4t - t^2 \;\Rightarrow\; \omega = 2t^2 - \frac{t^3}{3}$$
The direction reverses when $\omega = 0$ again: $2t^2 = \dfrac{t^3}{3} \Rightarrow t = 6$ s.
$$\theta = \int_0^6\left(2t^2 - \frac{t^3}{3}\right)dt = \left[\frac{2t^3}{3} - \frac{t^4}{12}\right]_0^6 = 144 - 108 = 36\ \text{rad}$$
Rotations $= \dfrac{36}{2\pi} \approx 5.7$: more than $3$ but less than $6$.
Solution by Sreeraj P, M.Sc Physics